An object flying in air with velocity $\left(20\hat{i} + 25\hat{j} - 12\hat{k}\right)$ suddenly breaks in two pieces whose masses are in the ratio $1 : 5$. The smaller mass flies off with a velocity $\left(100\hat{i} + 35\hat{j} + 8\hat{k}\right)$. The velocity of the larger piece will be :

Question

A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kg/s with a constant relative speed of 11 km/s. What is the acceleration of the rocket one minute after the blast?

Answer 1

To solve this problem, we need to apply the principle of conservation of momentum. The total momentum before and after the breakup of the object must be equal.

Let's denote:

Total initial velocity, $\vec{V}_i = 20\hat{i} + 25\hat{j} - 12\hat{k}$
Smaller mass's velocity after breakup, $\vec{V}_s = 100\hat{i} + 35\hat{j} + 8\hat{k}$
Larger mass's velocity after breakup, $\vec{V}_l$

The total mass of the original object is divided into two parts with a mass ratio of 1:5.

If the smaller mass is m, the larger mass will be 5m.

According to the conservation of momentum:

$ (m + 5m)\vec{V}_i = m\vec{V}_s + 5m\vec{V}_l $

Since the total mass cancels out from both sides, we can simplify to:

$\vec{V}_i = \frac{1}{6} \left(\vec{V}_s + 5\vec{V}_l\right) $

Substituting the known values, we get:

$(20\hat{i} + 25\hat{j} - 12\hat{k}) = \frac{1}{6}\left((100\hat{i} + 35\hat{j} + 8\hat{k}) + 5\vec{V}_l\right)$

Rearranging for $\vec{V}_l$:

$6(20\hat{i} + 25\hat{j} - 12\hat{k}) = (100\hat{i} + 35\hat{j} + 8\hat{k}) + 5\vec{V}_l$

$(120\hat{i} + 150\hat{j} - 72\hat{k}) - (100\hat{i} + 35\hat{j} + 8\hat{k}) = 5\vec{V}_l$

$20\hat{i} + 115\hat{j} - 80\hat{k} = 5\vec{V}_l$

Now, divide by 5 to solve for $\vec{V}_l$:

$\vec{V}_l = \frac{1}{5}(20\hat{i} + 115\hat{j} - 80\hat{k})$

$\vec{V}_l = 4\hat{i} + 23\hat{j} - 16\hat{k}$

Therefore, the velocity of the larger piece is $\boxed{4\hat{i} + 23\hat{j} - 16\hat{k}}$, which matches the correct option given.

Answer 2